I faced this question where it was already given that $\sum A_n$ is converging and I had to prove that $\sum (A_n - A_{n+1})$ is also convergent to the value $A_0$.

I proceeded assuming $A_n > A_{n+1}$ from $\lim [A_{n+1}/A_n] <1$ as it was given $\sum A_n$ is converging , so now if I were to expand $\sum (A_n - A_{n+1})$ starting from n = 0 till n then I would get $A_0 - A_1 + A_1 - A_2 +..........+ A_{n-1} - A_n$ henceforth in the end we get the result $A_0 - A_n$ due to cancellation.

Now my problem starts here that is how do I now prove $\sum (A_n - A_{n+1})$ is converging to $A_0$ ? can I say that $(A_0 - A_n) \to A_n$ since $A_n$ must be very small for the fact $\sum A_n$ is converging, so obviously if $A_n$ is a finite quantity it must be very much smaller as compared to $A_0$. Am I right in my approach or can anyone suggest a better method...?

  • $\begingroup$ "if An is a finite quantity it must be very much lesser than A0" Dunno what "much lesser than A0" means. But yes, if a series $\sum\limits_nx_n$ converges then $x_n\to0$, and this is all one needs here. $\endgroup$ – Did Oct 16 '15 at 6:59
  • $\begingroup$ Unrelated: you really have to work on your titles, the string of questions you posted so far have the most uninformative ones one can imagine. Done deal? $\endgroup$ – Did Oct 16 '15 at 7:01
  • $\begingroup$ @Did it won't be the case as you state if $x_n = 1/n$ $\endgroup$ – Arnav Das Oct 16 '15 at 7:02
  • $\begingroup$ Huh? You are confusing "If P then Q" with "If Q then P". $\endgroup$ – Did Oct 16 '15 at 7:03
  • 1
    $\begingroup$ Title for this question: done. Now you can change the titles of your former questions... $\endgroup$ – Did Oct 16 '15 at 7:05

We have

$$|A_0-\sum_0^M (A_{n}-A_{n+1})| = |A_0 - A_{M+1}| $$

Since $\sum A_n$ converges to say $A$ we have that for each $\epsilon>0$ there's a $L$ such that$

$$\forall M>L:|A-\sum_0^M A_n| < \epsilon$$

This means especially that $A_{M+1} < 2\epsilon$ which means that

$$\forall M>L:|A_0-\sum_0^M (A_{n}-A_{n+1})| = |A_0 - A_{M+1}| < 2\epsilon$$

  • $\begingroup$ One can prefer the OP's approach, computing the partial sums of the series for a direct conclusion. $\endgroup$ – Did Oct 16 '15 at 7:38
  • $\begingroup$ @Did Well, the initial approach is the same (my first expression), then the rest is to show that $A_0-A_{M+1}\to A_0$ which is the part he's missing. $\endgroup$ – skyking Oct 16 '15 at 8:25
  • $\begingroup$ But to show that $A_n\to0$ is much simpler than what you present. $\endgroup$ – Did Oct 16 '15 at 8:33
  • $\begingroup$ @Did How would you prove it in a much simplier way? The theorem that $x_n\to0$ if $\sum x_n$ converges is proved in pretty much the same way. Note that you can't use that theorem since the OP found it to be presumtious (ie something that can't be accepted without a proof). $\endgroup$ – skyking Oct 16 '15 at 8:38
  • $\begingroup$ For example, noting that if $(s_n)$ converges then $s_n-s_{n-1}\to0$, and using this for $s_n=\sum\limits_{k=1}^nx_k$. (Note that the OP seems to have consulted their notes since their first remark and to have come to the conclusion that this result was indeed in them.) $\endgroup$ – Did Oct 16 '15 at 8:42

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